The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.

Solution:

Let $\frac{a}{r}, a, a r$ be the first three terms of the G.P.

$\frac{a}{r}+a+a r=\frac{39}{10}$ $\ldots(1)$

$\left(\frac{a}{r}\right)(a)(a r)=1$ $\ldots(2)$

From (2), we obtain

$a^{3}=1$

$\Rightarrow a=1$ (Considering real roots only)

Substituting $a=1$ in equation (1), we obtain

$\frac{1}{r}+1+r=\frac{39}{10}$

$\Rightarrow 1+r+r^{2}=\frac{39}{10} r$

$\Rightarrow 10+10 r+10 r^{2}-39 r=0$

$\Rightarrow 10 r^{2}-29 r+10=0$

$\Rightarrow 10 r^{2}-25 r-4 r+10=0$

$\Rightarrow 5 r(2 r-5)-2(2 r-5)=0$

$\Rightarrow(5 r-2)(2 r-5)=0$

$\Rightarrow r=\frac{2}{5}$ or $\frac{5}{2}$

Thus, the three terms of G.P. are $\frac{5}{2}, 1$, and $\frac{2}{5}$.